IChO 1986 · Theory problem 7

Continuous process A + 2 B <=> 2 C: flow sheet and energy balance

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The equilibrium constant of the reaction \(\ce{A(g) + 2B(g) <=> 2C(g)}\) is \(K_p = 10.0\ \mathrm{MPa^{-1}}\). The starting materials are supplied at 25 °C and heated to 100 °C where complete equilibration takes place. Below 100 °C the reaction rate is negligibly small. The whole process is executed continuously in a stationary state. The boiling points at 0.1 MPa of \(\ce{A}\), \(\ce{B}\), and \(\ce{C}\) are 40 °C, 80 °C, and 60 °C, respectively. The three compounds have the same heat of evaporation: \(q\ \mathrm{J\,mol^{-1}}\). The heat capacities of \(\ce{A}\), \(\ce{B}\), and \(\ce{C}\) may be neglected. A schematic diagram of a distillation is shown below (Fig. 1).

Figure 1

The total heat used at each of the two distillations is \(3q\ \mathrm{J\,mol^{-1}}\) (of the top product). Apart from distillation columns (each with its own evaporator and condenser) the pieces of apparatus shown of Fig. 2 are available.

Figure 2

7.1 Draw a flow diagram of the process in which all flows are given (flow sheet) and in which the starting materials are used as efficiently as possible using as few pieces of apparatus as possible.

7.2 Express the equilibrium constant in terms of the degree of conversion and of total pressure for the case that the feed to the reactor is stoichiometric. What is the value of the degree of conversion when total pressure is 0.100 MPa?

7.3 Number all flows. Calculate the composition of each flow in \(\mathrm{mol\,s^{-1}}\) for a rate of production of 1 mole of \(\ce{C}\) per second under the conditions of part b.

7.4 In what respect can the reaction conditions influence the composition of the mixture that leaves the reactor? (Refer to question b.)

7.5 The process requires energy. For the conditions of b explain where energy should be supplied in the flow diagram drawn in part a. Derive an expression for the total energy needed.

Official English text of the problem.

Answers

  • 7.1: Combine fresh and recycled reactants in one evaporator, feed the vapor to the reactor, then separate the outlet in two columns: first take \(\ce{A}\) overhead; then take \(\ce{C}\) overhead from the remaining \(\ce{B}/\ce{C}\) mixture. Recycle all separated \(\ce{A}\) and \(\ce{B}\). A separate reactor-outlet condenser can be omitted if the first column accepts vapor feed.
  • 7.2: \[ \boxed{K_p=\frac{\alpha^2(3-\alpha)}{P(1-\alpha)^3}}, \qquad \boxed{\alpha=\frac13=33.3\%}. \]
  • 7.3: In \(\mathrm{mol\,s^{-1}}\), fresh feed is \(0.500\ \ce{A}+1.00\ \ce{B}\); recycle is \(1.00\ \ce{A}+2.00\ \ce{B}\); reactor feed is \(1.50\ \ce{A}+3.00\ \ce{B}\); reactor outlet is \(1.00\ \ce{A}+2.00\ \ce{B}+1.00\ \ce{C}\). The complete numbered stream table appears below.
  • 7.4: Higher pressure increases equilibrium conversion. The effect of temperature depends on the unspecified reaction enthalpy. Below 100 °C, the composition is effectively frozen by the negligible reaction rate.
  • 7.5: Vaporization requires \(4.50q\ \mathrm{J\,s^{-1}}\), and each column requires \(3.00q\ \mathrm{J\,s^{-1}}\): \[ \boxed{\dot Q_{\mathrm{vaporization+distillation}}=10.5q\ \mathrm{J\,s^{-1}}}. \] The additional signed reactor duty is \[ \boxed{\dot Q_R=(0.500\ \mathrm{mol\,s^{-1}})\Delta_{\mathrm r}H_g}, \] where \(\Delta_{\mathrm r}H_g\) is the enthalpy of the gas-phase reaction as written.

Solution

The symbols \(\ce{A}\), \(\ce{B}\), and \(\ce{C}\) denote unspecified compounds. Their chemical names and SMILES cannot be inferred from the data.

7.1 — Flow sheet

At 0.100 MPa, the volatility order indicated by the boiling points is

\[ \ce{A}\;(40\,^\circ\mathrm C) \;>\; \ce{C}\;(60\,^\circ\mathrm C) \;>\; \ce{B}\;(80\,^\circ\mathrm C). \]

Use the following arrangement, with the stream numbers also used in part 7.3:

  1. Join fresh liquid \(\ce{A}\), stream 1, and fresh liquid \(\ce{B}\), stream 2, with the two liquid recycle streams 6 and 9.
  2. Send the combined liquid, stream 3, through one common heater/evaporator. Its gaseous outlet, stream 4, enters the reactor at 100 °C.
  3. Send the equilibrium reactor outlet, stream 5, to the first distillation column.
  4. In the first column, obtain pure \(\ce{A}\) as the liquid top product, stream 6, and recycle it to the feed junction. Send the bottom product containing \(\ce{B}\) and \(\ce{C}\), stream 7, to the second column.
  5. In the second column, withdraw pure liquid \(\ce{C}\) as the top product, stream 8, and recycle pure liquid \(\ce{B}\), the bottom product, as stream 9.

Thus the pure liquid outlet temperatures are 40 °C for recycled \(\ce{A}\), 60 °C for product \(\ce{C}\), and 80 °C for recycled \(\ce{B}\).

Complete recycling gives complete overall utilization of the starting materials, despite incomplete conversion during each reactor passage. Two ordinary columns with top and bottom products provide the required separation of the intermediate-boiling product.

For the fewest separate units, stream 5 can enter the first column as vapor: its built-in condenser supplies the necessary condensation duty. If liquid feed to that column is required, insert one cooler/condenser between the reactor and the first column. Its inlet and outlet have identical molar compositions.

Removing \(\ce{A}\) first also gives the smaller distillation heating requirement, as shown in part 7.5.

7.2 — Equilibrium conversion

Let \(\alpha\) be the fraction of the entering \(\ce{A}\) converted. For a stoichiometric reactor feed initially containing \(n_0\) mol of \(\ce{A}\), \(2n_0\) mol of \(\ce{B}\), and no \(\ce{C}\),

\[ \ce{A(g) + 2B(g) <=> 2C(g)} \]

gives the equilibrium amounts

\[ n_A=n_0(1-\alpha),\qquad n_B=2n_0(1-\alpha),\qquad n_C=2n_0\alpha. \]

The total is

\[ n_{\mathrm{tot}}=n_0(3-\alpha). \]

Consequently,

\[ p_A=P\frac{1-\alpha}{3-\alpha},\qquad p_B=P\frac{2(1-\alpha)}{3-\alpha},\qquad p_C=P\frac{2\alpha}{3-\alpha}. \]

Substitution into the equilibrium expression gives

\[ K_p=\frac{p_C^2}{p_Ap_B^2} =\boxed{\frac{\alpha^2(3-\alpha)} {P(1-\alpha)^3}}. \]

At \(P=0.100\ \mathrm{MPa}\),

\[ K_pP=(10.0\ \mathrm{MPa^{-1}})(0.100\ \mathrm{MPa})=1.00. \]

Therefore,

\[ \alpha^2(3-\alpha)=(1-\alpha)^3, \]

\[ 3\alpha^2-\alpha^3 =1-3\alpha+3\alpha^2-\alpha^3, \]

and hence

\[ \boxed{\alpha=\frac13=33.3\%}. \]

The equilibrium mole fractions are \(y_A=1/4\), \(y_B=1/2\), and \(y_C=1/4\). As a check,

\[ K_p= \frac{(0.0250)^2}{(0.0250)(0.0500)^2} =10.0\ \mathrm{MPa^{-1}}. \]

7.3 — Numbered material balances

Producing \(1.00\ \mathrm{mol\,s^{-1}}\) of \(\ce{C}\) requires a reaction extent rate

\[ \dot\xi=\frac{\dot n_C}{2} =0.500\ \mathrm{mol\,s^{-1}}. \]

Since one-third of the entering \(\ce{A}\) reacts,

\[ \dot n_{A,\mathrm{in}} =\frac{\dot\xi}{\alpha} =\frac{0.500}{1/3} =1.50\ \mathrm{mol\,s^{-1}}, \]

and the stoichiometric feed contains

\[ \dot n_{B,\mathrm{in}}=3.00\ \mathrm{mol\,s^{-1}}. \]

The reactor therefore discharges

\[ \dot n_{A,\mathrm{out}}=1.50-0.500=1.00, \]

\[ \dot n_{B,\mathrm{out}}=3.00-2(0.500)=2.00, \]

\[ \dot n_{C,\mathrm{out}}=2(0.500)=1.00, \]

all in \(\mathrm{mol\,s^{-1}}\).

The complete external stream table is:

Stream Description \(\ce{A}\), \(\mathrm{mol\,s^{-1}}\) \(\ce{B}\), \(\mathrm{mol\,s^{-1}}\) \(\ce{C}\), \(\mathrm{mol\,s^{-1}}\)
1 Fresh liquid \(\ce{A}\) 0.500 0 0
2 Fresh liquid \(\ce{B}\) 0 1.00 0
3 Combined liquid feed to evaporator 1.50 3.00 0
4 Gaseous feed to reactor 1.50 3.00 0
5 Reactor outlet to first column 1.00 2.00 1.00
6 First-column top product; \(\ce{A}\) recycle 1.00 0 0
7 First-column bottom product to second column 0 2.00 1.00
8 Second-column top product; final \(\ce{C}\) 0 0 1.00
9 Second-column bottom product; \(\ce{B}\) recycle 0 2.00 0

The fresh feed exactly replaces the reactants consumed. Thus the overall material balance corresponds to

\[ \ce{A + 2B -> 2C}, \]

with no reactants discarded.

7.4 — Influence of reaction conditions

For the specified stoichiometric feed,

\[ K_pP=\frac{\alpha^2(3-\alpha)}{(1-\alpha)^3}. \]

The right-hand side increases with \(\alpha\) over \(0<\alpha<1\). Therefore, at fixed temperature, increasing pressure increases conversion. This agrees with the reduction from three gaseous reactant molecules to two gaseous product molecules.

Temperature changes \(K_p\). The van ’t Hoff relation is

\[ \frac{d\ln K_p}{dT} =\frac{\Delta_{\mathrm r}H_g}{RT^2}. \]

Accordingly:

  • If the forward reaction is endothermic, increasing temperature increases equilibrium conversion.
  • If it is exothermic, decreasing temperature increases equilibrium conversion.

The reaction enthalpy is not supplied, so the applicable direction cannot be determined. Moreover, below 100 °C the reaction is negligibly slow: cooling the outlet preserves its composition during separation.

Changing the reactant ratio also changes the equilibrium composition; an excess of one reactant generally increases the fractional conversion of the other. A catalyst or additional residence time cannot change the final equilibrium composition once equilibrium has been reached.

7.5 — Energy requirements

Let the molar latent heat be

\[ L=q\ \mathrm{J\,mol^{-1}}. \]

Because heat capacities are neglected, there is no sensible-heating contribution from raising the feeds to 100 °C.

The common evaporator vaporizes all fresh and recycled reactants:

\[ \dot Q_E =(1.50+3.00)\ \mathrm{mol\,s^{-1}}\,L =\boxed{4.50q\ \mathrm{J\,s^{-1}}}. \]

The first column produces \(1.00\ \mathrm{mol\,s^{-1}}\) of overhead \(\ce{A}\), so its reboiler requires

\[ \dot Q_{D1} =(1.00\ \mathrm{mol\,s^{-1}})(3L) =\boxed{3.00q\ \mathrm{J\,s^{-1}}}. \]

The second column produces \(1.00\ \mathrm{mol\,s^{-1}}\) of overhead \(\ce{C}\):

\[ \dot Q_{D2} =\boxed{3.00q\ \mathrm{J\,s^{-1}}}. \]

Thus the specified vaporization and distillation heating duties total

\[ \boxed{ \dot Q_{\mathrm{physical}} =(4.50+3.00+3.00)q =10.5q\ \mathrm{J\,s^{-1}} }. \]

Equivalently, these operations require \(10.5q\ \mathrm J\) per mole of \(\ce{C}\).

The alternative separation order—first taking \(\ce{A}+\ce{C}\) overhead, then separating them—would have overhead rates of \(2.00\) and \(1.00\ \mathrm{mol\,s^{-1}}\). Its distillation duties would total \(9.00q\), compared with \(6.00q\ \mathrm{J\,s^{-1}}\) for the chosen order.

The reactor must also be maintained at 100 °C. Define \(\Delta_{\mathrm r}H_g\) as the molar enthalpy of

\[ \ce{A(g) + 2B(g) -> 2C(g)}. \]

Its signed heat requirement is

\[ \boxed{ \dot Q_R =\dot\xi\,\Delta_{\mathrm r}H_g =(0.500\ \mathrm{mol\,s^{-1}})\Delta_{\mathrm r}H_g }. \]

The algebraic sum of the evaporator, reboiler, and reactor duties is therefore

\[ \boxed{ \dot Q_E+\dot Q_{D1}+\dot Q_{D2}+\dot Q_R = 10.5q\ \mathrm{J\,s^{-1}} +(0.500\ \mathrm{mol\,s^{-1}})\Delta_{\mathrm r}H_g }. \]

An endothermic reaction requires additional heat at the reactor; an exothermic reaction requires heat removal there. With no heat recovery, the total positive heating demand is

\[ \boxed{ \dot Q_{\mathrm{heating}} = 10.5q\ \mathrm{J\,s^{-1}} +\max\!\left[0,\, (0.500\ \mathrm{mol\,s^{-1}})\Delta_{\mathrm r}H_g \right] }. \]

For an energy-balance check, the reactor outlet contains \(4.00\ \mathrm{mol\,s^{-1}}\) of vapor. In the vapor-fed layout, the first column’s condenser rejects \(7.00q\ \mathrm{J\,s^{-1}}\): \(4.00q\) from condensation of that feed plus its \(3.00q\) reboiler input. The second condenser rejects \(3.00q\ \mathrm{J\,s^{-1}}\). A separate effluent condenser would divide the first \(7.00q\) into \(4.00q+3.00q\), leaving the total unchanged.

After subtracting condenser cooling, the overall net heat exchange is consequently

\[ \dot Q_{\mathrm{net}} = 0.500q\ \mathrm{J\,s^{-1}} +(0.500\ \mathrm{mol\,s^{-1}})\Delta_{\mathrm r}H_g. \]

The solution was checked against the official solution.